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2. exercise

ASEN 3112 Structures - Fall 2014: Homework 4
Solutions
Exercise 4-1
A solid circular shaft of 40 mm diameter is to be replaced by a hollow circular tube made of the same material. The
outside diameter of the hollow tube is limited to 60 mm. Find: (a) the thickness of the hollow tube if both designs
are to work at the same maximum shear stress, (b) the ratio of weights between the two shafts, commenting on
which one is more weight efficient. Use the exact theory for circular shafts presented in Lecture 7. Hint: solve a
quartic equation.
Solution. Let T be the given torque, while Rs and Rt are the external radii of the solid and tube shafts,
respectively. (Note: the exact theory for torsion of circular shafts is used because the hollow tube that
solves the problem is not necessarily thin since its thickness is unknown; that way one covers all bases.)
The corresponding polar moment of inertias are Js = π Rs4 /2 and Jt = π [Rt4 − (Rt − t)4 ]/2, where t is
the internal tube thickness. Equating the maximum shear stress in both shafts gives the condition
τmax =
T Rs
T Rs
T Rt
T Rt
= Js = 1
=
= 1 4
4
Js
Jt
π Rs
π Rt − (Rt − t)4
2
2
Cancelling out T and 12 π, inserting Rs = 40/2 = 20 mm and Rt = 60/2 = 30 mm, and rationalizing
yields the equation
20 × 304 − (30 − t)4 = 30 × 204
Solving gives (30 − t)4 = 570, 000 mm4 , which has 4 roots for 30 − t, two real. The real root giving
positive thickness is 30 − t = 27.477 mm, or
t = 2.523 mm
The weight ratio of solid shaft to tube is the ratio of their cross section areas:
As
π Rs2
202
Ws
=
=
= 2.758
=
Wt
At
302 − 27.4772
π[Rt2 − (Rt − t)2 ]
Hence the tube weights 2.758 times less than the solid shaft.
The cross sections of the original (solid) and replacement (hollow) shaft are picture below to scale.
(Drawing these figures is not required in the HW solution; they are shown here to illustrate the fact that
the hollow tube uses less material.)
Since the hollow shaft is indeed thin wall, the CTW approximate theory could have been used, and the
answer would be very close. However, that possibility was not known a priori.
Replacement shaft
Original shaft
1
Exercise 4-2
10 kW
off
(a)
25 kW
off
40 kW
x
(b)
−1325 Nm
E
E
−530 Nm
D
5 kW
Bearing off
A
6.0 m
C
B
D
−265 Nm
A
6.0 m
B
3.0 m
1.5 m
m
C
+2120 Nm (motor drive)
The solid 50-mm-diameter steel (G = 84 GPa) line shaft shown in Figure 4.1(a) is driven by a 40-kW motor at 3
Hz. The power is transmitted to gears located at B, D and E, which take up power as shown. Converting power to
torque† gives the N-m values listed in Figure 1(b). (For the applied torque signs, note that the longitudinal x axis is
directed from A to E.)
(a) Find the maximum shear stresses in sections BC, C D and D E of the shaft. (Draw first a torque diagram that
plots T along the shaft from A through E. In between gears T is constant.)
(b) Determine the total angle of twist in degrees between A and E. Note: signs of internal torques are important for
this computation.
Solution. (a) The drive power acting on the bearings at B, C, D and E given in kW at a driving frequency
of f = 3 Hz, is converted to torque in Nm using . The results of the conversion are shown in (b) of the
figure above, already provided in the HW assignment. The sense adopted for the driven torque is arbitrary
because the only important thing for max stress is magnitude. However if the motor-applied torque at C,
which is 40 kW × 159 Nm/(kW.Hz)/3 Hz = 2120 Nm, is taken as positive, the other three, absorbed by
the gears at B, D and E must be obviously negative. Equilibrium check: −265+2120−1325−530 = 0.
Using these values and using the method of sections, the internal torque that acts on each segment is
easily obtained by inspection. The values are shown in the following diagram:
Internal torques in each segment
0
0
A
−35 Kw
−1855 Nm
5 Kw
265 Nm
B
−5 kW
−265 Nm
−10 Kw
−530 Nm
C
40 kW
+2120 Nm
D
−25 kW
−1325 Nm
E
−10 kW
−530 Nm
Applied torques at B, C, D, E
† P = 2π f T , where P is the power and f is the rotational frequency in Hz. In SI units, 1 watt (W) = 1 Nm/sec.
2
The polar moment of inertia of the shaft is J = 12 πr 4 = 12 π0.0254 = 6.14 × 10−7 m4 . The maximum
shear stresses in the shaft segments, ignoring signs, are
τ AB = 0 since
TBC r
τ BC =
=
J
TC D r
=
τC D =
J
TD E r
=
τD E =
J
T AB = 0,
265 × 0.025
N
= 10.8 × 106 2 = 10.8 MPa
−7
6.14 × 10
m
N
1855 × 0.025
= 75.6 × 106 2 = 75.6 MPa
−7
6.14 × 10
m
N
530 × 0.025
= 21.6 × 106 2 = 21.6 MPa
−7
6.14 × 10
m
Hence the maximum shear stress occurs over segment C D.
(b) From equation (6-15) of text, the angle of twist between A and E is the sum of the twist angles of
the individual segments because the internal torque is constant over each one:
φ AE = φ AB + φ BC + φC D + φ D E =
T AB L AB
TBC L BC
TC D L C D
TD E L D E
+
+
+
GJ
GJ
GJ
GJ
in which the signs of the internal torques T AB , TBC , etc, are now important. Substituting values:
φ AB = 0,
−265 Nm × 3 m
= −0.0154 rad,
6.14 × 10−7 m4 × 84 × 109 N/m2
1855 Nm × 6 m
=
= 0.2158 rad,
6.14 × 10−7 m4 × 84 × 109 N/m2
530 Nm × 6 m
=
= 0.0616 rad,
6.14 × 10−7 m4 × 84 × 109 N/m2
φ BC =
φC D
φD E
whence
φ AE = 0 − 0.0154 + 0.2158 + 0.0616 = 0.262 rad = 15.01◦
Note on sign conventions. The exercise was taken from the Mechanics of Materials textbook by E. P.
Popov. To agree with Vable’s sign convention for torque, the x axis should be directed from A to E. If
so the twist angle is −15◦ instead of 15◦ . Credit should be given if the answer differs only in sign; since
the only important value is the magnitude.
3
Exercise 4-3
20
m
m
4 mm throughout
20 mm
100 mm
The figure above shows the cross section of a simplified wing torque box. The thickness t = 4 mm is uniform. For
items (a)–(c) carry out the analysis using CTW theory, whereas for (d) use OTW theory.
(a)
Determine the torque T = Tmax that can be carried at a maximum allowable shear stress of 20 MPa (this value
already incorporates a safety factor). Use midline dimensions for the calculations.
(b)
Suppose the thickness of the two inclined plates is increased while keeping midlines the same. Will the
maximum torque capacity increase?
(c)
Compute the torsional rigidity G J in N-m2 if the wingbox material is Al 6061-T6, for which G = 25.9 GPa.
Since t is constant, to get J it is OK to use equation (9.5) of posted Lecture 9.
(d)
The torque box is slit by a longitudinal cut of negligible width along the back fin. It is submitted to the
maximum torque Tmax found in (a). Using OTW theory with α = β = 1/3, find the ratios of the max shear
stress and torsional rigidities, respectively, with respect to the values found above for the CTW case. Comment
on how much the cut has weakened the torque box: would one made of Al-6061-T6 (yield shear stress ≈ 138
MPa) fail?
Solution.
(a) The midline enclosed area is
) = 760 + 880 + 2016 = 3656 mm2 = 3656 × 10−6 m2
A E = ( 12 π 222 + 44 × 20 + 12 44 × (100 22
24
Using T = 2A E q = 2A E τmax t, the maximum torque is found to be
T = 2 × 3656 × 10−6 m2 × 20 × 106 N/m2 × 0.004 m = 585 Nm
Note: the centerline horizontal dimension of the triangular area: 100(22/24) = 91.67 mm, is approximate. If one uses 100 mm, A E becomes 3840 × 10−6 m2 and T increases to 644 Nm.
(b) Because the shear flow is constant and τ = q/t, increasing the thickness of the inclined plates would
reduce τ in those plates and help reduce stress concentration at the sharp trailing edge corner. But it
would not affect the other portions of the profile.
(c) We have
4A2
J= E
ds
t
where the enclosed area A E has been computed above. We need the contour integral
1
1
ds
= π × 22 + 2 × 20 + 2 222 + 91.672 = (69 + 40 + 188.5) = 74.4,
t
4
4
4
which is a dimensionless number (the expression in parentheses is the midline perimeter). The shear
modulus for Al 6061-T6 is G = 25.9 GPa. Hence the torsional rigidity of the section is
2
4 × 3656 × 10−6 m2
GJ =
× 25.9 × 109 N/m2 = 18612 N-m2
74.4
Note: if the centerline horizontal dimension of the triangular area is taken as 100 mm, the rigidity G J
increases to 19470 N-m2 .
(d) Sketchy since it was done by Mathematica (see code below) and transcribed here; students should
show more work.
It follows that cutting the box would increase the maximum shear stress roughly 18.5 times , which
will surely cause failure in Al-6061-T6 if the torque found in item (a) is applied.
Solution of torque-box problem, item (d), using Mathematica:
Solution for Exercise 4.3, CTW case
ClearAll[m,mm,Nw,MPa];
m=1000*mm; MPa=10^6*Nw/m^2; t=4*mm; τ maxCTW=20.0*MPa;
JCTW=2*((N[Pi]/2)*22^2+20*44+22*(100*22/24))*t*τ maxCTW;
Print["JCTW=",Simplify[JCTW]];
JCTW =
585109. Nw
mm
Solution for Exercise 4.3, OTW case
ClearAll[m,mm,Nw,MPa];
m=1000*mm; MPa=10^6*Nw/m^2;
JCTW=585109.*Nw/mm; Print["JCTW=",JCTW]; (* from cell above *)
TCTW=579.2*Nw*m; τ maxCTW=20.0*MPa; Print["τ maxCTW=",τ maxCTW];
Rm=22.0*mm; a=20.0*mm; d=90.0*mm; t=4.0*mm;
b=Rm*N[Pi]+2*a+2*Sqrt[a^2+d^2]; b=Simplify[b,mm>0]; Print["b=",b];
JOTW=Simplify[(1/3)*b*t^3,mm>0];
Print["JOTW=",JOTW,"
JCTW/JOTW= ",Simplify[JCTW/JOTW,mm>0]];
τ maxOTW=Simplify[TCTW*t/JOTW];
Print["τ maxOTW=",τ maxOTW,
"
τ maxOTW/τ maxCTW= ",Simplify[τ maxOTW/τ maxCTW,mm>0]];
585109. Nw
mm
20. Nw
τ maxCTW =
mm2
b=293.506 mm
JCTW =
JOTW=6261.46 mm4
τ maxOTW =
370.01 Nw
mm2
JCTW/JOTW =
93.4461 Nw
mm5
τ maxOTW /τ maxCTW = 18.5005
5
Exercise 4-4.
bb = 2 in
tb = 1 in
tp =p 1/2 in
dp = 4 in
Upon getting your diploma you find a job at Whirlpool and come up with the design of the above figure for the
agitator shaft in a washing machine prototype. This is fabricated by welding four rectangular blades to a circular
pipe of the dimensions shown. Each blade is 1 by 2 in. The material of the pipe and blades is the same. If the
maximum shear stress, neglecting stress concentrations, is limited to 10 ksi, what torque T = Tmax (units: in-kips)
can be applied to the shaft? (Additional hint material in assignment omitted for brevity)
Solution. Following the hints in the assignment assume that the total torque T is expressed as
T = Tp + 4Tb ,
(1)
in which Tp is taken by the pipe and Tb by each of the blades. Let Jp and Jb denote the torsional constants
of the pipe and of each of the blades, respectively.
π
π
Jp = (re4 −ri4 ) = (24 −1.54 ) = 17.204 in4 , Jb = β bb tb3 = 0.229 × 2 × 13 = 0.458 in4 . (2)
2
2
Here Jp is the polar moment of inertia of the pipe section, as befits an annular cross section treated by
the exact theory of Lecture 7, while Jb = Jbβ = β bb tb3 comes from the OTW theory of Lecture 8. The
value of β = 0.229 is extracted from the table in (8.3) of Lecture 8 for the aspect ratio bb /tb = 2. (The
approximation β ≈ 1/3 is way off in this problem since the blade sections are not thin rectangles.)
The total J for the agitator assembly is J = Jp + 4Jb = 19.036 in4 . Since the value of G is the same
for pipe and blade, the proportion of total torque taken by each of the components is given by the ratios
Jp
Jb
T = 0.9036 T, Tb =
T = 0.0241 T.
(3)
J
J
(numerical check: 0.9036 + 4 × 0.0241 = 1.) The max shear stresses in the pipe and each blade are
Tp =
Tp r e
0.9036 T × 2
τallow
10
=
= 0.1051 T ⇒ Tpmax = max =
= 95.1 kips-in
Jp
17.204
τp
0.1051
Tb tb
0.0241 T × 1
τallow
10
=
=
= 0.0407 T ⇒ Tbmax = max =
= 245.6 kips-in
3
Jbα
0.246 × 2 × 1
τb
0.0407
τ pmax =
τbmax
(4)
Here the expression for τbmax = T tb /Jbα , with Jbα = α bb tb3 , comes from the OTW theory, with
α = 0.246 read off from the table in Lecture 8 for bb /tb = 2. From (4) we conclude that the agitator
torque capacity is limited by the pipe:
Tmax = min (95.1, 245.6) = 95.1 kips-in
6